Well the co-efficent of x^2 is given solely by a on the left and d on the right, for the two equations to be equal this co-efficent must be the same therfore a=d again the same with b and e for the co-efficent of x and also c and f must be equal as they give the units (or the co-efficent of x^0 if you like).

Originally posted by Hurkyl Form three equations by plugging in -1, 0, and 1 for x. Solve those equations for a, b, and c in terms of d, e, and f.

How do you know that the numbers -1, 0, 1 will not give a special case result ?
i mean, i know if you plug in -1, 0, 1 you will get a=d, b=e, c=f, but how you would explain that this is not a special case (you know, you chose three special numbers, so the result can be special case result).

We're told that the equation holds for all x, so it must be true for any choices of x!

By solving the equations, we've proven that, if the equation is true for all x, then a = d, b = e, and c = f. It's trivial to go the other way and prove that if a = d, b = e, and c = f then the equation is true for all x.

No, they are not "equal by definition". The definition of
"f(x)= g(x)" is that for any value x= a, the values f(a) and g(a) are equal. While the value, of course, depends on the coefficients, it is incorrect to say that the definition of equality of two polynomials is that they have the same coefficients.

I would also note that jcsd's suggestion (which is essentially the same thing) works, it probably would not be accepted as a proof since I suspect that the purpose of this problem is to see WHY the fact that two polynomials have the same values for all x implies that they have the same coefficients.

f(x) can be factored as such because we know f(x) has n different roots, which means (x - xj) must be a factor of x for each root xj. Multiplying out all n of these factors makes it clear we must also have an additional factor an.

Originally posted by Hurkyl Choose n to be the highest exponent of x with a nonzero coefficient.

f(x) can be factored as such because we know f(x) has n different roots, which means (x - xj) must be a factor of x for each root xj. Multiplying out all n of these factors makes it clear we must also have an additional factor an.

Okay, here is my point. You stated that if the highest coefficient is non-zero, then there are precisely n roots of f(x) (counting "double", "triple" roots their respective number of time). If we assume f(x) has n + 1 distinct roots, then we immediately know the highest coefficient must be zero for this situation to even possibly exist. Unfortunantly, because the highest coefficient is now zero, we don't know whether we can rewrite the polynomial as desired. Therefore, we must avoid taking that route in a proof.

Although from here the proof is relatively easy to complete. Now that the highest coefficient is zero we have a new equivalent polynomial of degree n-1 which equals f(x) (but we don't know whether this new polynomial's highest coefficient is zero). Again, since we have n+1 distinct roots to f(x), and our law states that a polynomial of degree n-1 can have at most n-1 distinct roots if it's highest coefficient is non-zero, then we know (since this is not the case) that the new equivalent polynomial of degree n-1 has a highest coefficient of zero. Repeating this for all coefficients completes the proof.

There are lots of ways to actually prove this. Though I think you and Kam have essentially the same proof, just Kam left out a detail or two. My mental proof is slightly different, and I would employ the "let m be the highest integer so that am is inequal to 0)" approach.

Originally posted by Hurkyl There are lots of ways to actually prove this. Though I think you and Kam have essentially the same proof, just Kam left out a detail or two. My mental proof is slightly different, and I would employ the "let m be the highest integer so that am is inequal to 0)" approach.

I think there is a bit of confusion. I was only referring to KL Kam's lemma, where the goal was to prove that all the coefficients are zero (i.e. there is no highest non-zero coefficient). Also, I must disagree that his lemma only requires a quick fix as that transformation (which he cannot preform) is the key step. This is certainly not a crusade against KL Kam's work. I respect his response and only feel it justified to critique considering the time he put in. Anyways, enough of that, we all know that it is provable and there is no need to spend more time on it.